Question

Consider the standard BCS theory but assume that the interaction energy U

that enters the definition of gap parameter (i.e. Δ=(U/N)∑ₖ⟨c_ₖ↓cₖ↑⟩
, where N
is the number of sites) is large compared to the Fermi energy (BEC regime). The zero-temperature gap equation in this limit can be simply solved to give Δ(0)=U/2
, so I would expect that the critical temperature scales accordingly T_c ≈ Δ (0) ≈ U
. However, we know that mean field theory does not capture the Goldstone mode (corresponding to fluctuations of the phase), hence I would expect that, as long as the superfluid is charge-neutral, there are low lying excitations of the phase field that cost less energy than the energy required to break a pair and so I would expect a much lower critical temperature.

First question: is this idea correct? Can I intuitively guess how T_c
scales with U
in this case? My guess, based on some literature, is T_c ≈ 1 / U
because the superfluid stiffnes (or superfluid density) at zero temperature goes as 1 / U
.

Now comes the real question: how is the picture above modified for a charged superfluid, where the minimal coupling to a gauge field must be included in the theory to ensure gauge invariance? In this case there is the Anderson-Higgs mechanism, i.e. the gauge field becomes massive, but it is not very clear to me what happens to the Goldstone mode... I guess it simply does not exist anymore? So in this case should I expect Tc ≈ U
as if no Goldstone mode exists whatsoever?

More details In case my question is not clear enough, I would also appreciate a description of the spectral function beyond mean-field BCS theory. In the latter case, there is just a spectral gap that increases with U
. If we include Goldstone mode on a neutral superfluid, the gap is filled by energy states corresponding to collective modes (is this right? can you sketch how?). How is the spectral function modified when the superfluid is charged? Bonus: I guess that the single-particle spectral function −1 / π T r I m G ( ω + i 0⁺ ), where G is the Green function, cannot capture the collective modes, so what should I calculate to see the collective modes?

Any help, even partial answers, are very appreciated, thanks!

Answer 1

In the BCS theory under BEC regime, Tc is thought to scale inversely with the interaction energy U. For charged superfluids, the Anderson-Higgs mechanism alters this relationship, potentially leading Tc to scale similarly to U. The spectral function in such superconductors is modified beyond mean-field approximation to include collective excitations.

Step-by-step explanation:

In the BCS theory of superconductivity, the critical temperature (Tc) is crucial for understanding when a material will exhibit superconducting properties. The consideration of the interaction energy (U) being large compared to the Fermi energy, particularly in the Bose-Einstein Condensate (BEC) regime, leads to an intuitive guess that Tc scales inversely with U. This is because the superfluid stiffness or density, which supports the flow of superconducting pairs without resistance, tends to scale as 1/U.

When considering a charged superfluid, the picture changes significantly due to the Anderson-Higgs mechanism, where the gauge field gains mass and the Goldstone mode, associated with phase fluctuations in a charge-neutral superfluid, is essentially 'eaten' by the gauge field to become a longitudinal component of the massive gauge boson. This may suggest a return to a scaling of Tc similar to U as in the simple BCS case. Spectral functions beyond mean-field theory often involve including contributions from collective excitations (like Goldstone modes in neutral superfluids), which fill in the spectral gap. In charged superconductors, these are affected by the coupling to the gauge field. To see collective modes in the spectral function, one might need to compute two-particle Green's functions or use other many-body techniques that can capture these excitations.